Which type of wave displaces the particles of the medium parallel to the direction on which the wave travels?

Answer:

C

Explanation:

The men bow to the women and the women curtsy to the men.

(took on test and got it right)

Answer:

Gravity

Explanation:

Gravity pulls things down to earth and it is a force

Answer:

False

Explanation:

No, it is not true that energy of the electrons ejected should have been proportional to the intensity of the light. Perhaps the kinetic energy of the ejected electrons is independent of the intensity of the incident radiation. This is the very fact that classical theory of electromagnetism fails to explain in photoelectric effect. The kinetic energy of the electrons remains constant even if the amplitude of the incident light is increased.

Answer:

1) m = 0.3137 kg

2a)T_f = -181.7°C

2b) T_f = -1176.97°C

Explanation:

1) We are given;

Length; l = 4.30 cm = 0.043 m

Width; w = 14.0 cm = 0.014 m

height; h = 27.0 cm = 0.027 m

density of gold; ρ = 19.3 × 10³ kg/m³

Formula for the density is known as;

ρ = mass/volume

Thus;

m =ρV

m = 19.3 × 10³ × (lwh)

m = 19.3 × 10³ × (0.043 × 0.014 × 0.027)

m = 0.3137 kg

2a) We are given;

Diameter of brass; L_br = 10 cm

Diameter of aluminum; L_al = 10.01 cm

Now, to some for change in temperature we will use the formula;

L_f = L_i + αL_i(Δt)

Where α is coefficient of expansion.

Now, for the ring to be removed from the rod, the final diameter of the brass has to be same as the aluminium.

Thus;

L_f(brass) = L_f(aluminium)

From table attached, α_brass ≈ 19 × 10^(-6) /°C

Also, α_aluminium ≈ 24 × 10^(-6) /°C

Thus;

L_f(brass) = 10 + (19 × 10^(-6) × 10 × (Δt))

Similarly,

L_f(aluminium) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Since L_f(brass) = L_f(aluminium), then;

10 + (19 × 10^(-6) × 10 × (Δt)) = 10.01 + (24 × 10^(-6) × 10.01 × (Δt))

Rearranging, we have;

10.01 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.01 = Δt(-50.24 × 10^(-6))

Δt = 0.01/(-50.24 × 10^(-6))

Δt ≈ -199°C

Thus, temperature at which the combination must be cooled to separate the two metals is;

T_f = T_i + Δt

T_f = 17.3 + (-199)

T_f = -181.7°C

2b) Diameter of aluminum is now;

L_al = 10.06 cm

Thus;

10.06 - 10 = (19 × 10^(-6) × 10 × (Δt)) - (24 × 10^(-6) × 10.01 × (Δt))

0.06 = Δt(-50.24 × 10^(-6))

Δt = 0.06/(-50.24 × 10^(-6))

Δt = -1194.27°C

T_f = 17.3 + (-1194.27)

T_f = -1176.97°C

Answer:

v = -17.94 cm

Explanation:

Given that,

The candle is placed at a distance of 50 cm, u = -50 cm

The focal length of the diverging lens, f = -28 cm (negative in case of a concave lens)

We need to find the image distance. We know that the lens formula is as follows:

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(-28)}+\dfrac{1}{(-50)}\\\\v=-17.94\ cm[/tex]

So, the image distance is equal to 17.94 cm.

Answer:

By talking about something that interesto you’

sorry if wrong

Explanation:

Answer; they are larger and made of rocky material

Answer:

C

Explanation:

C

Answer:

It will go up to 93.75 m before it is moving at 20 m/s

Explanation:

As we know that

[tex]v^2 - u^2 = 2aS[/tex]

here v is the final speed i.e 20 m/s

u is the initial speed i.e 5 m/s

a is the acceleration due to gravity i.e 2 m/s^2

Substituting the given values in above equation, we get -

[tex]20^2 - 5^2 = 2*2*S\\S = 93.75[/tex]meters

im confused hold on imma send you a link to the answerExplanation:

Answer:

It will bob 7.887640449 times a minute

Explanation:

I hope this is correct!!

Answer:

The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].

Explanation:

The work done is given as in terms of

[tex]W=\Delta TE[/tex]

Where ΔTE is the change in total energy.

This is given as

[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]

Rearranging it in  terms of K_x gives

[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]

Answer:

[tex]0.0035\ \text{m}[/tex]

Explanation:

y = Distance from the center point

d = Separation between slits = 0.3 mm

D = Distance between slit and screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = 700 nm

m = Order = 1

We have the relation

[tex]d\dfrac{y}{D}=m\lambda\\\Rightarrow y=\dfrac{Dm\lambda}{d}\\\Rightarrow y=\dfrac{1.5\times 1\times 700\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y=0.0035\ \text{m}[/tex]

The distance from the screen at which the first bright fringe beyond the center fringe appear is [tex]0.0035\ \text{m}[/tex].

Answer:

get egg and try to make in not crack when it falls by exerting the momentum of the fall into something other than the egg ex. make a box full of bubble wrap and put your egg in it

Explanation:

Answer:

huh

Explanation:

Answer:

B : using a spring scale and measuring the weight of the car

Explanation:

Answer:

[tex]0.3165\ \text{rad/s}[/tex]

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = [tex]\dfrac{d}{2}=3.25\ \text{m}[/tex]

v = Velocity of person running = 3.8 m/s

[tex]I_t[/tex] = Moment of inertia of turntable = [tex]1850\ \text{kg m}^2[/tex]

Moment of inertia of the system is

[tex]I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2[/tex]

As the angular momentum of the system is conserved we have

[tex]L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}[/tex]

The angular velocity of the turntable is [tex]0.3165\ \text{rad/s}[/tex].

Answer:

[tex]0.0018833\ \text{m/s}[/tex]

Explanation:

[tex]d[/tex] = Distance of Andromeda Galaxy from Earth = [tex]2.54\times 10^7\ \text{ly}[/tex]

[tex]t[/tex] = Time taken = [tex]90\ \text{years}[/tex]

[tex]c[/tex] = Speed of light = [tex]3\times 10^8\ \text{m/s}[/tex]

We have the relation

[tex]t=t_o\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow 90=2.54\times 10^7\sqrt{1-\dfrac{v^2}{c^2}}\\\Rightarrow \dfrac{90^2}{(2.54\times 10^7)^2}=1-\dfrac{v^2}{c^2}\\\Rightarrow 1-\dfrac{90^2}{(2.54\times 10^7)^2}=\dfrac{v^2}{c^2}\\\Rightarrow v=c\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}}[/tex]

[tex]c-v=c(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=3\times 10^8(1-\sqrt{1-\dfrac{90^2}{(2.54\times 10^7)^2}})\\\Rightarrow c-v=0.0018833\ \text{m/s}[/tex]

The required answer is [tex]0.0018833\ \text{m/s}[/tex].

Answer:

the car movves briefly as you ran, however, it stops again after you ran in to the wall

Explanation:

Our net total linear momentum was zero at the time both the train and the boy was at rest. As there is no pressure, the train and boy system's linear momentum can be conserved provided no external forces are working on it that might shift its momentum.

If the boy runs inside of the train with a velocity V1 in the forward direction, the train would have a velocity V2 in the reverse direction to V1 to conserve the system's momentum, resulting in Final momentum.

i.e

[tex]m \times V_1 + M \times V_2 = 0[/tex]  --- (1)

here;

m = mass of the boy

M = mass of the train

Thus;

[tex]m \times V_1 =- M \times V_2[/tex] --- (2)

As the boy crashes into the train's wall, a pair of equal and opposing force F intervene on both the train and the boy. This force F causes the boy traveling with momentum m× V1 to come to a halt; its momentum remains zero. As the train moves with about the same momentum as the boy as seen in (2) and is subjected to the same force F, its momentum will be diminished to zero, and it will come to a halt.

Answer: 84 %

Explanation:

Answer:

The box will not move from its position.

Explanation:

First, we will calculate the static frictional force that is stopping the box to move from its position:

[tex]f = \mu R = \mu W=\mu mg[/tex]

where,

f = static frictional force = ?

μ = coefficient of static friction = 0.3

m = mass of box = 4 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]f = (0.3)(4\ kg)(9.81\ m/s^2)\\f=11.77\ N[/tex]

Since the frictional force (11.77 N) is greater than the applied force (10 N).

Therefore, the box will not move from its position.

Answer:

Waves carry energy from one place to another.

Explanation:

Because waves carry energy, some waves are used for communication, eg radio and television waves and mobile telephone signals.

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

a). 1.28333 seconds

b). 186.66 seconds

Explanation:

a). Given :

Distance between the earth and the moon, d = [tex]$3.85 \times 10^8$[/tex] m

Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{3.85 \times 10^8}{3 \times 10^8}$[/tex]

 = 1.28333 seconds

b). Distance between Mars and the earth, d = [tex]$5.6 \times 10^{10}$[/tex] m

   Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

So, the time required for his voice to reach earth is :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$[/tex]

 = 186.66 seconds

Answer:

a)   y = 3.73 m,  b)  t = 1.74 s,  c)  R = 40.99 m,

d) vₓ  = 23.49  m/s,   v_y = -8.5 m / s

Explanation:

This is a projectile launching exercise, we start by breaking down the initial velocity

          sin θ = v_{oy} / v₀

          cos θ = v₀ₓ / v₀

          v_{oy} = v₀ sin θ

          v₀ₓ = v₀ cos θ

          v_{oy} = 25 sin 20 = 8.55 m / s

          v₀ₓ = 25 cos 20 = 23.49 m / s

a) when the egg reaches the maximum height its vertical speed is zero

          v_y² = v_{oy}² - 2 g y

          0 = v_[oy}² - 2g y

           y = v_{oy}² / 2g

          y = [tex]\frac{8.55^2}{2 \ 9.8 }[/tex]

          y = 3.73 m

b) flight time

          y = v_{oy} t - ½ g t²

the time of flight occurs when the body reaches the ground y = 0

          0 = (v_{oy} - ½ g t) t

The results are

          t₁ = 0s        this time is for using the body star

          v_{oy} - ½ g t = 0

           t = [tex]\frac{2v_{oy}^2}{g}[/tex]

           t = 2 8.55 / 9.8

           t = 1.74 s

c) the range

           R = v₀² sin 2θ / g

           R = 25² sin (2 20) / 9.8

           R = 40.99 m

d) speed at the point of arrival

horizontal speed is constant

           vₓ = v₀ₓ = 23.49  m/s

vertical speed is

           v_y = Iv_{oy} - g t

           v_y = 8.55 - 9.8  1.74

           v_y = -8.5 m / s

Explanation:

The answer is In the picture. Thanks.

hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.

Hope this helps!

Answer:

Potential energy and kinetic energy are constituents of mechanical energy.

When a turbine is switched on, it rotates with mechanical energy.

Since a motor runs the turbine, it converts this mechanical energy to electrical energy.